1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Epsilon-limit proof for real number sequences

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Let {an} be a sequence of real numbers.
    Suppose an->L as n->∞. Prove that [(a1+a2+...+an)/n] ->L as n->∞.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    By definition:
    an->L iff
    for all ε>0, there exists an integer N such that n≥N => |an - L|< ε.

    Given ε>0.
    I start with |[(a1+a2+...+an)/n] - L|. But I cannot think of a way to link this with |an - L| so I have no idea how to continue.

    Any help is appreciated! :)
     
  2. jcsd
  3. Jan 18, 2010 #2
    I think we may have to use the triangle inequality, but still I have no idea how to get into the exact the same form as |an - L|.

    Could someone help me, please?
     
    Last edited: Jan 18, 2010
  4. Jan 18, 2010 #3
    I don't get this part.

    In the hypothesis, we suppose an->L, so:
    for all εa>0, there exists an integer N such that n≥N => |an - L|< εa.

    So for example:
    Given ε>0. We can find N s.t. n≥N => |an-L|< ε/4

    But how come you have |(a1+...+an) / n|?

    Thanks for your help! :)
     
  5. Jan 18, 2010 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It isn't. I must have been in a fog when I posted that and I have deleted it. Be back a little later.
     
  6. Jan 18, 2010 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Given [itex]\varepsilon > 0[/itex] pick N such that if n > N then [itex]|a_n - L| < \varepsilon/ 4[/itex]. Now if n > N

    [tex]\left|\frac{a_1+...+a_n}{n}-L\right| = \left|\frac{a_1+...+a_n}{n}-\frac {nL}{n}\right|= \left|\frac{(a_1-L)+...+(a_n-L)} n\right|[/tex]

    [tex]= \left| \frac{(a_1-L)+...+(a_N-L)}{n}+\frac{(a_{N+1}-L)+...+(a_n-L)}{n}\right|[/tex]

    Now use the triangle inequality. Since N is fixed you can make the first term small in n is large enough. Then work on the second term to make it small. Sorry for the false start earlier.
     
  7. Jan 18, 2010 #6
    1) hmm...why did you choose ε/4 in [itex]|a_n - L| < \varepsilon/ 4[/itex]??

    2) At the end, we have to show that:
    for all ε>0, there exists an integer M such that
    n≥M => |[(a1+a2+...+an)/n] - L|< ε.
    But how can we find such an M?

    Thanks for your help! :)
     
    Last edited: Jan 18, 2010
  8. Jan 18, 2010 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Because I anticipate adding a couple of small terms together and having the sum come out less than [itex]\varepsilon[/itex].

    Have you done the triangle inequality on this yet, so you have two terms to make small? If you can say there exists an N1 such that n > N1 makes the first term small and an N2 such that n > N2 makes the second term small, then if you take Nmax = max{N1,N2,N}, wouldn't that do it? Of course, you have to fill in the arguments why you can pick these N's.
     
  9. Jan 18, 2010 #8
    So after using the triangle inequality (which will break it into 2 terms), for the first term I have to find N1 such that n≥N1 => |[(a1 -L)+...(aN -L)]/n| < ε/2

    To find this N1, I think I have to relate |[(a1 -L)+...(aN -L)]/n| to |an - L|, but how is this possible?? I can't think of a way of doing it...

    Thanks!
     
  10. Jan 18, 2010 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    N is a fixed, finite, number. There are finitely many terms in the numerator and you have n in the denominator. Just pick n big enough...
     
  11. Jan 18, 2010 #10
    I get the intuitive idea, but just like every epsilon-limit proof of this type, we have to actually find a specific N1 which works, right? And I'm stuck on this step...
     
  12. Jan 18, 2010 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So solve your inequality |[(a1 -L)+...(aN -L)]/n| < ε/2 for n to see how large it must be.
     
  13. Jan 18, 2010 #12
    OK. I think it is n> 2 |(a1-L)+...+(aN-L)| / ε.
    So I think we can take N1 = 2 |(a1-L)+...+(aN-L)| / ε, is this correct?
    But on top of N depending on ε, N also depends on the terms in the sequence, is this OK? (i.e. is this allowed?)
     
  14. Jan 18, 2010 #13

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Everything depends on the terms of the sequence. You wouldn't expect the same N to work for all sequences, would you?

    Remember you might have to take N1 slightly larger than that because that might not be an integer.
     
  15. Jan 18, 2010 #14
    So I guess I will take N1 to be the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.


    Any hints for the second term? (sorry I'm asking it but I really have no clue...)

    |[aN+1 - L +...+ (an-1-L) +(an-L)] / n|

    Here there are terms like an, an-1, etc. The "n" pops up everyhwere, so I don't think I can possibly solve for n in the inequality just like I did before.
     
  16. Jan 18, 2010 #15
    The sequence an converges therefore it is bounded. So every an<M where M is some number. So the sum becomes less that (M-L) ...n-N+1 times over n and from there go on.
     
  17. Jan 18, 2010 #16
    |[aN+1 - L +...+ (an-1-L) +(an-L)] / n|
    ≤|(n-N)(M-L)/n|

    I think it should be n-N instead of n-N+1?
    n-(N+1)+1 = n-N
     
  18. Jan 18, 2010 #17

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Somewhere in this proof you need to use the fact that the original sequence converges to L. Remember at the beginning where you had that if n > N,

    [itex]
    |a_n - L| < \varepsilon/ 4
    [/itex]

    Use that now.
     
  19. Jan 19, 2010 #18
    Do I have to use the fact that any convergence sequence is bounded?

    But how should I handle the terms like (an-1-L) in the numerator?

    thanks.
     
  20. Jan 19, 2010 #19

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I fear you are getting lost in the subscripts. In the expression:

    |[aN+1 - L +...+ (an-1-L) +(an-L)] / n|

    all the subscripts are greater than N. So every one of them satisfies the inequality:

    [itex]

    |a_k - L| < \varepsilon/ 4

    [/itex]

    for whatever value from N+1 to n that k takes. Use that.
     
  21. Jan 19, 2010 #20
    So should I break |[(aN+1 - L) +...+ (an-1-L) +(an-L)] / n| using the triangle inequality into n-N terms?
    |[(aN+1 - L) +...+ (an-1-L) +(an-L)] / n| <= (1/n)[|aN+1 - L| + ...+ |an-1-L| + |an-L|] ?

    But I don't think it will work...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook